In a "Rotor-ride" at a carnival, people are rotated in a cylindrically walled "room." The radius of the room is 5.5 m, The coefficient of static friction is μ=0.1.
What is the minimum rotation frequency such that the people will not slide down the wall when the floor drops out?
If it goes halfway around in one second, then the period of the ride must be 2.0 seconds.
What is going on is that the ride is pushing on the people to make them go in a circle. For every action there is an equal but opposite reaction, and so the people push back on the ride. The centripetal force the ride exerts on people becomes the normal force that causes the friction that keeps them from sliding down the walls, and that friction as such must be larger than their weight, or down the wall they go. There is nothing pushing them to the outside, rather they are accelerating toward the middle of the circle. Remember that you feel "thrown" the opposite direction of the acceleration. (i.e. if your car accelerated forward, you are thrown backward)
So, the normal force is just the centripetal force acting on the people: (I pick the centripetal force with the period in it)
FN = m4ð2r/T2
Since the friction force must be larger than or equal to the weight of the riders, the minimum coefficient of static friction will be:
mg = Ffr = µFN = µ(m4ð2r/T2)
And if you plug the centripetal force in for the normal force, you get:
mg = µ(m4ð2r/T2) - so let's solve for µ:
g = µ(4ð2r/T2) (canceling the mass)
gT2/(4ð2r) = µ (divide both sides by 4ð2r/T2)
µ = gT2/(4ð2r) = (9.80 m/s/s)(2.0 s)2/[4ð2(5.0 m)] = .1986 = .20
To determine the minimum rotation frequency to prevent people from sliding down the wall, we need to consider the forces acting on them.
When the floor drops out, the people will experience a centrifugal force due to the circular motion. This force is directed outward and can be calculated using the formula:
Fc = m * ω^2 * r
Where:
Fc is the centrifugal force
m is the mass of a person
ω is the angular velocity (in radians per second)
r is the radius of the room
The frictional force between the people and the wall opposes the centrifugal force. The maximum static friction force can be calculated using the formula:
Fs = μ * N
Where:
Fs is the maximum static friction force
μ is the coefficient of static friction
N is the normal force
In this case, the normal force is equal to the weight of the person. So we have:
N = m * g
Where:
g is the acceleration due to gravity
Since the maximum static friction force should be equal to or greater than the centrifugal force, we can set up the following equation:
μ * N ≥ Fc
Substituting the values and rearranging the equation, we get:
μ * m * g ≥ m * ω^2 * r
The mass of the person cancels out, and we are left with:
μ * g ≥ ω^2 * r
Rearranging the equation to solve for ω, we get:
ω ≥ sqrt(μ * g / r)
Substituting the given values of μ, g, and r, we can calculate the minimum rotation frequency:
ω ≥ sqrt(0.1 * 9.8 / 5.5)
ω ≥ sqrt(0.098 / 5.5)
ω ≥ sqrt(0.0178)
ω ≥ 0.133 radians per second
So, the minimum rotation frequency required to prevent people from sliding down the wall is approximately 0.133 radians per second.